The Unapologetic Mathematician

Mathematics for the interested outsider

Distinct Eigenvalues

Today I’d like to point out a little fact that applies over any field (not just the algebraically-closed ones). Let T be a linear endomorphism on a vector space V, and for 1\leq i\leq n, let v_i be eigenvectors with corresponding eigenvalues \lambda_i. Further, assume that \lambda_i\neq\lambda_i for i\neq j. I claim that the v_i are linearly independent.

Suppose the collection is linearly dependent. Then for some k we have a linear relation

\displaystyle v_k=c_1v_1+c_2v_2+...+c_{k-1}v_{k-1}

We can assume that k is the smallest index so that we get such a relation involving only smaller indices.

Hit both sides of this equation by T, and use the eigenvalue properties to find

\displaystyle\lambda_kv_k=c_1\lambda_1v_1+c_2\lambda_2v_2+...+c_{k-1}\lambda_{k-1}v_{k-1}

On the other hand, we could just multiply the first equation by \lambda_k to get

\displaystyle\lambda_kv_k=c_1\lambda_kv_1+c_2\lambda_kv_2+...+c_{k-1}\lambda_kv_{k-1}

Subtracting, we find the equation

\displaystyle0=c_1(\lambda_k-\lambda_1)v_1+c_2(\lambda_k-\lambda_2)v_2+...+c_{k-1}(\lambda_k-\lambda_{k-1})v_{k-1}

But we this would contradict the minimality of k we assumed before. Thus there can be no such linear relation, and eigenvectors corresponding to distinct eigenvalues are linearly independent.

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February 4, 2009 - Posted by | Algebra, Linear Algebra

5 Comments »

  1. [...] vectors here, they span a subspace of dimension , which must be all of . And we know, by an earlier lemma that a collection of eigenvectors corresponding to distinct eigenvalues must be linearly [...]

    Pingback by Transformations with All Eigenvalues Distinct « The Unapologetic Mathematician | February 10, 2009 | Reply

  2. [...] Eigenspaces are Disjoint We know that eigenspaces of distinct eigenvalues are disjoint. That is, no vector is an eigenvector of two different eigenvalues. This extends to generalized [...]

    Pingback by Generalized Eigenspaces are Disjoint « The Unapologetic Mathematician | February 25, 2009 | Reply

  3. [...] as a corollary we can find that not only are the eigenvectors corresponding to distinct eigenvalues linearly independent, they are actually orthogonal! Indeed, if and are eigenvectors of with [...]

    Pingback by Eigenvalues and Eigenvectors of Normal Transformations « The Unapologetic Mathematician | August 6, 2009 | Reply

  4. Is the dimension of V, n? ie dim(V)=n, ie their contains n elements in a basis for V. since you have n eigenvectors that are linearly independent and any set of n linearly independent vectors is a basis for an n dimensional vectorspace.

    Comment by H | April 18, 2010 | Reply

  5. Yes, H, that’s true. But all this post was intended to prove was their linear independence.

    Oh, and I think you mean “there”.

    Comment by John Armstrong | April 19, 2010 | Reply


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